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Ques. When I am going East at 10 km per hour a train moving with constant velocity appears to be moving exactly north east. When my velocity is increased to 30 kmph east it appears to be moving north. Q1 what is the velocity of train along north in kmph ? Q2 what is the velocity of train along east in kmph? Q3 at what speed in kmph should I move north so that train appears to be moving exactly south east ?

Ques. When I am going East at 10 km per hour a train moving with constant velocity appears to be moving  exactly north east. When my velocity is increased to 30 kmph east it appears to be moving north.
Q1 what is the velocity of train along north in kmph ?
Q2 what is the velocity of train along east in kmph?
Q3 at what speed in kmph should I move north so that train appears to be moving exactly south east ?

Grade:11

2 Answers

Susmita
425 Points
5 years ago
Let the train have a velocity
\vec{v} = a i+bj
When you are moving east with speed 10kmph your velocity
\vec{u} = 10i
Relative velocity of train wrt you is
\vec{r} = \vec{v}-\vec{u}=(a-10)i+bj
But as the train appears to move exactly north east so  b/(a-10)=tan45
Or,a-10=b
Now you are moving east with a speed 30kmph.Now the relative velocity of train
\vec{r`} = \vec{v}-\vec{u}=(a-30)i+bj
But now the train appears to be moving north.
So a-30=0
So a=30
So b=20
Trains velocity=30i+20j
Along east its velocity is 30kmph and along north is 20kmph.
Now let you move with a velocity uj along north.
Relative velocity of train wrt you is
\vec{r} = \vec{v}-\vec{u}=30i+(10-u)j
But you see the train south east.
so tan(90+45)=(10-u)/30
Or,-1=(10-u)/30
u=40kmph
 
 
 
Nitin Kumar Oberoi
13 Points
one year ago
Let the train have a velocity
\vec{v} = a i+bj
When you are moving east with speed 10kmph your velocity
\vec{u} = 10i
Relative velocity of train wrt you is
\vec{r} = \vec{v}-\vec{u}=(a-10)i+bj
But as the train appears to move exactly north east so b/(a-10)=tan45
Or,a-10=b
Now you are moving east with a speed 30kmph.Now the relative velocity of train
\vec{r`} = \vec{v}-\vec{u}=(a-30)i+bj
But now the train appears to be moving north.
So a-30=0
So a=30
So b=20
Trains velocity=30i+20j
Along east its velocity is 30kmph and along north is 20kmph.
Now let you move with a velocity uj along north.
Relative velocity of train wrt you is
\vec{r} = \vec{v}-\vec{u}=30i+(20-u)j
But you see the train south east.
so tan(90+45)=(20-u)/30
Or,-1=(20-u)/30
u=50kmph

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